Animated Solution for Physics - Physics and Measurement: In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a periodic motion is T=2π5g7(R−r). The values of R and r are measured to be (60±1) mm and (10±1) mm, respectively. In five successive measurements, the time period is found to be 0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59 s. The least count of the watch used for the measurement of time period is 0.01 s. Which of the following statement(s) is (are) true?
Select Answer:
* Multiple Correct
Visualized Solution
Physical Setup
T=2π5g7(R−r)
This is the time period of a solid sphere rolling inside a hollow sphere.
Mean Time Period (Tmean)
Tmean=50.52+0.56+0.57+0.54+0.59
Tmean=52.78=0.556 s
Rounding off to 2 decimal places:
Tmean≈0.56 s
Mean Absolute Error (ΔTmean)
ΔTi=∣Tmean−Ti∣
ΔT1=∣0.56−0.52∣=0.04 s
ΔT2=∣0.56−0.56∣=0.00 s
ΔT3=∣0.56−0.57∣=0.01 s
ΔT4=∣0.56−0.54∣=0.02 s
ΔT5=∣0.56−0.59∣=0.03 s
ΔTmean=50.04+0.00+0.01+0.02+0.03=0.02 s
% Error in T
% error in T=TmeanΔTmean×100
% error in T=0.560.02×100
% error in T≈3.57%
Option (b) is correct.
% Error in r
r=10±1 mm
% error in r=rΔr×100
% error in r=101×100=10%
Option (a) is correct.
Error Equation for g
T=2π5g7(R−r)
T2=5g4π2⋅7(R−r)
g=5T228π2(R−r)
gΔg×100=R−rΔ(R−r)×100+2TΔT×100
% Error in (R−r)
R=60±1 mm,r=10±1 mm
R−r=60−10=50 mm
Δ(R−r)=ΔR+Δr=1+1=2 mm
% error in (R−r)=502×100=4%
Final % Error in g
gΔg×100=4%+2(3.57%)
gΔg×100=4%+7.14%
gΔg×100≈11%
Option (d) is correct.
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The Sigma Insight: Errors in Measurement
Solution Diagram
The problem of error analysis often feels like a puzzle where every small uncertainty cascades into a larger one. In this JEE Advanced 2016 question, we are given an experiment to determine the acceleration due to gravity, g, using a periodic motion setup. The formula provided is T=2π5g7(R−r).
While the formula might look intimidating, it actually represents the time period of a solid sphere rolling inside a hollow sphere. However, the beauty of error analysis is that we don't need to derive the formula; we just need to dissect it! Let's break down the problem step by step.
Analyzing the Time Period
First, we need to find the most accurate value for the time period, T, from the five successive measurements: 0.52 s, 0.56 s, 0.57 s, 0.54 s, and 0.59 s.
The mean time period is simply the average of these values:
Tmean=50.52+0.56+0.57+0.54+0.59=0.556 s
Since our raw data is precise up to two decimal places, we must respect the rules of significant figures and round off our mean to two decimal places. Thus, Tmean≈0.56 s.
Next, we calculate the absolute error for each reading by finding the difference between the mean value and the individual readings:
ΔT1=∣0.56−0.52∣=0.04 s
ΔT2=∣0.56−0.56∣=0.00 s
ΔT3=∣0.56−0.57∣=0.01 s
ΔT4=∣0.56−0.54∣=0.02 s
ΔT5=∣0.56−0.59∣=0.03 s
The mean absolute error is the average of these errors:
ΔTmean=50.04+0.00+0.01+0.02+0.03=0.02 s
Now, we can find the percentage error in the time period:
% error in T=TmeanΔTmean×100=0.560.02×100≈3.57%
This confirms that option (b) is correct.
Error in the Radius
The problem gives the measurement of the small radius as r=10±1 mm. The percentage error is straightforward:
% error in r=101×100=10%
This makes option (a) correct as well.
The Master Equation for Gravity
To find the error in g, we need to rearrange our master equation to make g the subject. Squaring both sides of T=2π5g7(R−r), we get:
T2=5g4π2⋅7(R−r)
g=5T228π2(R−r)
In error analysis, constants like 28π2 and 5 have zero uncertainty. The maximum percentage error in g is the sum of the percentage errors of the variables involved. Since T is squared, its percentage error is multiplied by 2:
gΔg×100=R−rΔ(R−r)×100+2TΔT×100
Final Calculation
Let's evaluate the error in the term (R−r). We are given R=60±1 mm and r=10±1 mm.
The value of (R−r) is 60−10=50 mm.
Crucially, when subtracting quantities, their absolute errors always add up to account for the worst-case scenario. So, Δ(R−r)=1+1=2 mm.
The percentage error in (R−r) is:
% error in (R−r)=502×100=4%
Finally, we substitute everything back into our error equation for g:
gΔg×100=4%+2(3.57%)=4%+7.14%≈11%
This confirms that option (d) is also correct.
By systematically breaking down the measurements and applying the rules of error propagation, we've successfully navigated through the uncertainties to find the correct statements!
